Shift down the elements one step and add a diagonal of ones in the main diagonal.

Change the sign of all the elements below the main diagonal. is found as a comment here: Oeis table A121207.

Calculate the matrix inverse and you will get:

As we see we have the Bell numbers in the first column. Oeis, Bell numbers. The INVERT transform of any number triangle is equivalent to the steps above, here with the INVERT transform of the Pascal triangle as example.

Repeating the procedure or algorithm (above), infinitely many times, will produce the Catalan numbers as a convergent Oeis Catalan numbers in all columns, regardless of starting triangle (or starting sequence). That is you input matrix into the first step instead of matrix .

Another way to calculate the first column in matrix A is by taking matrix powers of this matrix M. Here the elements in the main diagonal have been deleted except for the first element which is equal to 1. Taking matrix powers is simply matrix multiplication repeated, , and so on.

As promised we have calculated the Bell numbers in the first column.

Yet another way to calculate the first column in matrix , is to calculate permanents of a modified version of matrix . Here the element in the lower right corner has been swapped with the element in the lower right corner.

Hi! This is really neat. I could really use the iterated invert function, but I’m not sure I’m doing it right! When I put a starting series
(S =x +x^2+x^3+…)
into (1/(1-S))-1 it gives the expected results. However, no amount of iterating by putting those results back in for S seems to converge to the catalan numbers. Hints?

Wow, thanks! The algorithm is clear. But I’m going to have to study the relationship between the matrices and the generating functions. The link you gave at the top explained the INVERT transform in terms of generating functions and it seemed to correspond exactly to the matrices. However, maybe there is a big difference between INVERT on the generating function of a sequence, and INVERT on a (triangular) matrix. (I gather that the trinagular matrix corresponding to that sequence should have colomns with the sub diagonal entries the terms of the sequence as in Cat above.)

Ok–now I think I got it. Rather than the prescription for generating functions given in the link above (they say to take S(x) to (1/(1-S(x))) -1) I tried what the matrix algorithm actually said and took S(x) to 1/(1-x*S(x)). This indeed seems to give the Catalan number after iterating. Cool. I think there must be a species theory proof of this.